In the figure, a 6.40 g bullet is fired into a 0.725 kg block attached to the en

Question

In the figure, a 6.40 g bullet is fired into a 0.725 kg block attached to the end of a 0.310 m nonuniform rod of mass 0.332 kg. The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about A is 0.0895 kg-m2. Treat the block as a particle. (a) What then is the rotational inertia of the block-rod-bullet system about point A? (b) If the angular speed of the system about A just after impact is 9.43 rad/s, what is the bullet's speed just before impact? Rod Block Bulle (a) Number Units (b) Number Units

Explanation / Answer

Here, for the system

moment of inertia about A

a) moment of inertia = 0.0064 * 0.310^2 + 0.0895 + 0.725 * 0.310^2

moment of inertia = 0.1598 Kg.m^2

the moment of inertia is 0.1598 Kg.m^2

b)

let the initial speed of the bullet is v

Using conservation of angular momentum

0.0064 * 0.310 * v = 0.1598 * 9.43

solving for v

v = 760 m/s

the speed of the bullet before collision is 760 m/s

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